Let X1 and X2 be two independent exponentially distributed

Let X₁ and X₂ be two independent exponentially distributed random variables with means 0.5 and 0.25 respectively. Then Y = min (X₁, X₂) is

1. exponentially distributed with mean 1/6
2. exponentially distributed with mean 2
3. normally distributed with mean 3/4
4. normally distributed with mean 1/6

Mean (X₁) = 0.5

	1	= 0.5
	λ₁

λ₁ =	1	= 2
	0.5

Mean (X₂) = 0.25

	1	= 0.25 ⇒ λ₂ = 4
	λ₂

y= mean (X₁, X₂)

Mean (y) =	1	=	1	=	1
	y₁ + y₂		2 + 4		6